//
//  Problem1476.swift
//  TestProject
//
//  Created by 毕武侠 on 2021/5/4.
//  Copyright © 2021 zhulong. All rights reserved.
//

import UIKit

/*
 1476. 子矩形查询
 请你实现一个类 SubrectangleQueries ，它的构造函数的参数是一个 rows x cols 的矩形（这里用整数矩阵表示），并支持以下两种操作：

 1. updateSubrectangle(int row1, int col1, int row2, int col2, int newValue)

 用 newValue 更新以 (row1,col1) 为左上角且以 (row2,col2) 为右下角的子矩形。
 2. getValue(int row, int col)

 返回矩形中坐标 (row,col) 的当前值。
  

 示例 1：
 输入：
 ["SubrectangleQueries","getValue","updateSubrectangle","getValue","getValue","updateSubrectangle","getValue","getValue"]
 [[[[1,2,1],[4,3,4],[3,2,1],[1,1,1]]],[0,2],[0,0,3,2,5],[0,2],[3,1],[3,0,3,2,10],[3,1],[0,2]]
 输出：
 [null,1,null,5,5,null,10,5]
 解释：
 SubrectangleQueries subrectangleQueries = new SubrectangleQueries([[1,2,1],[4,3,4],[3,2,1],[1,1,1]]);
 // 初始的 (4x3) 矩形如下：
 // 1 2 1
 // 4 3 4
 // 3 2 1
 // 1 1 1
 subrectangleQueries.getValue(0, 2); // 返回 1
 subrectangleQueries.updateSubrectangle(0, 0, 3, 2, 5);
 // 此次更新后矩形变为：
 // 5 5 5
 // 5 5 5
 // 5 5 5
 // 5 5 5
 subrectangleQueries.getValue(0, 2); // 返回 5
 subrectangleQueries.getValue(3, 1); // 返回 5
 subrectangleQueries.updateSubrectangle(3, 0, 3, 2, 10);
 // 此次更新后矩形变为：
 // 5   5   5
 // 5   5   5
 // 5   5   5
 // 10  10  10
 subrectangleQueries.getValue(3, 1); // 返回 10
 subrectangleQueries.getValue(0, 2); // 返回 5
 示例 2：

 输入：
 ["SubrectangleQueries","getValue","updateSubrectangle","getValue","getValue","updateSubrectangle","getValue"]
 [[[[1,1,1],[2,2,2],[3,3,3]]],[0,0],[0,0,2,2,100],[0,0],[2,2],[1,1,2,2,20],[2,2]]
 输出：
 [null,1,null,100,100,null,20]
 解释：
 SubrectangleQueries subrectangleQueries = new SubrectangleQueries([[1,1,1],[2,2,2],[3,3,3]]);
 subrectangleQueries.getValue(0, 0); // 返回 1
 subrectangleQueries.updateSubrectangle(0, 0, 2, 2, 100);
 subrectangleQueries.getValue(0, 0); // 返回 100
 subrectangleQueries.getValue(2, 2); // 返回 100
 subrectangleQueries.updateSubrectangle(1, 1, 2, 2, 20);
 subrectangleQueries.getValue(2, 2); // 返回 20
  

 提示：

 最多有 500 次updateSubrectangle 和 getValue 操作。
 1 <= rows, cols <= 100
 rows == rectangle.length
 cols == rectangle[i].length
 0 <= row1 <= row2 < rows
 0 <= col1 <= col2 < cols
 1 <= newValue, rectangle[i][j] <= 10^9
 0 <= row < rows
 0 <= col < cols

 */
@objcMembers class Problem1476: NSObject {
    func solution() {
        let s = SubrectangleQueriesHistory([[1,1,1],[2,2,2],[3,3,3]])
        print(s.getValue(0, 0)); // 返回 1
        s.updateSubrectangle(0, 0, 2, 2, 100);
        print(s.getValue(0, 0)); // 返回 100
        print(s.getValue(2, 2)); // 返回 100
        s.updateSubrectangle(1, 1, 2, 2, 20);
        print(s.getValue(2, 2)); // 返回 20
    }
    
    /*
     暴力破解
     */
    class SubrectangleQueries {
        var list:[[Int]] = []
        
        init(_ rectangle: [[Int]]) {
            list = rectangle
        }
        
        func updateSubrectangle(_ row1: Int, _ col1: Int, _ row2: Int, _ col2: Int, _ newValue: Int) {
            for row in row1...row2 {
                for col in col1...col2 {
                    list[row][col] = newValue
                }
            }
        }
        
        func getValue(_ row: Int, _ col: Int) -> Int {
            return list[row][col]
        }
    }
    
    /*
     历史记录发
     1: 创建一个数组history将每次更改的记录下来
     2: 查找时，从history的后面开始遍历，看看是否这个位置在更改的范围内
     */
    
    class SubrectangleQueriesHistory {
        var list:[[Int]] = []
        var historys: [[Int]] = []
        
        init(_ rectangle: [[Int]]) {
            list = rectangle
        }
        
        func updateSubrectangle(_ row1: Int, _ col1: Int, _ row2: Int, _ col2: Int, _ newValue: Int) {
            historys.append([row1, col1, row2, col2, newValue])
        }
        
        func getValue(_ row: Int, _ col: Int) -> Int {
            for i in (0..<historys.count).reversed() {
                if row >= historys[i][0] && row <= historys[i][2] && col >= historys[i][1] && col <= historys[i][3] {
                    return historys[i][4]
                }
            }
            return list[row][col]
        }
    }
}
